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1. Laxmikant D. Rangari, Mrs. P.M. Zode, P.G. Mehar / International Journal of Engineering
Research and Applications (IJERA) ISSN: 2248-9622 www.ijera.com
Vol. 2, Issue4, July-August 2012, pp.2278-2281
Stress Analysis Of Lpg Cylinder Using Ansys Software
Laxmikant D. Rangari1 Mrs. P.M. Zode2 P.G. Mehar3
Student M Tech. Assistant Professor, Assistant Professor,
K.D.K.C.E., K.D.K.C.E., K.D.K.C.E.,
NAGPUR, NAGPUR, NAGPUR,
440009 440009 440009
ABSTRACT:
Computer aided investigations are Stress analysis of LPG gas cylinder IS 3196 by
carried using ANSYS, to verify maximum stress using finite element method.
and its location. To predict detailed stress 3D The cylinder is subjected to an internal pressure of
solid model has been chosen with the help of 2.5 MPa .Type material used is low carbon steel
PROE software. Two different types of non- .The material properties are :
linear FE models, plane and shell, were Young's Modulus 2.e+005 MPa
developed using 2D axisymmetric finite plane Poisson's Ratio 0.3
and shell elements, respectively. To create these Density 7.85e-006 kg/mm³
FE models and simulate the experimental burst, Tensile Yield Strength 240. MPa
first, shell MPs and thickness variations of the
LPG tanks due to spinning processes are Compressive Yield
240. MPa
investigated and input to the computer modeling Strength
processes. Additionally, after selecting the Tensile Ultimate Strength 420. MPa
loading and boundary conditions and
appropriate finite elements, the nonlinear ANSYS PROCEDURE FOR F.E. ANALYSIS
axisymmetric 2D FE models were generated and  Model
simulated in non-uniform and non-homogeneous o Geometry- Imported from PROE in “.iges”
conditions. [2] format
1. Solid- generated ansys geometry.
Keywords: LPG, Detailed stress, IS 3196, FEA o Mesh- tetrahedral element selection
modeling.  Static Structural
1. Analysis Settings- analysis settings are used for
INTRODUCTION static structural, single step loading
The pressure cylinder models assumed a 2. Loads-
cylinder with both ends capped. To simplify  Internal pressure of 2.5 Mpa is given on internal
modeling and hand calculations, both ends were wall of cylinder
treated as part of the cylinder. The hand calculations  The base of cylinder is kept fixed
from Roark’s Formulas for Strain table 28, 3. Solution-
condition 1c were used for comparison.  Total Deformation
[4]  Equivalent Stresses
 shear stress along xy plane
 Material Data
 low carbon steel
[3]
FIG.NO.1 LPG gas cylinder IS3196
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2. Laxmikant D. Rangari, Mrs. P.M. Zode, P.G. Mehar / International Journal of Engineering
Research and Applications (IJERA) ISSN: 2248-9622 www.ijera.com
Vol. 2, Issue4, July-August 2012, pp.2278-2281
PROE model Pressure of 2.5 Mpa is given on the internal surface
of the cylinder as shown in fig.
ANSYS11 .iges from PROE
The fixed support is given on the bottom of the
cylinder.
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3. Laxmikant D. Rangari, Mrs. P.M. Zode, P.G. Mehar / International Journal of Engineering
Research and Applications (IJERA) ISSN: 2248-9622 www.ijera.com
Vol. 2, Issue4, July-August 2012, pp.2278-2281
Mesh- mesh with automatic mesh settings is
generated.
The maximum equivalent stresses are
found ranging from 0 to 156 MPa. As shown in
figure, the contour plot of von mises stresses shows
the distribution of the stresses are concentrated
along the valve area of the cylinder and the bottom
portion.
The shearing effect is negligible as
compared to equivalent stresses. i.e 27 MPa. That
Observing the following results, the total can be found concentrated at the bottom of the
deformation in the cylinder for the internal stresses cylinder.
of 2.5 MPa is found to be maximum at the top i.e.
highlighted in red, and the minimum is found at the
fixed position i.e shown in blue colour.
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4. Laxmikant D. Rangari, Mrs. P.M. Zode, P.G. Mehar / International Journal of Engineering
Research and Applications (IJERA) ISSN: 2248-9622 www.ijera.com
Vol. 2, Issue4, July-August 2012, pp.2278-2281
The burst pressure of LPG gas cylinder has
been determined by use of finite element analysis
max. and min. equivalent stress, maximum shear
stress and deformation at critical area has been
calculated . The result of FEA analysis by using
ANSYS software are verified with hand
calculation. For comparison with ANSYS Software
result hand calculation has been done.
CONCLUSION:
The finite element simulation employed to
estimate the stresses of LPG gas cylinder via
ANSYS 11 software. The critical location of burst
pressure can be investigated by using ANSYS
software which is difficult by analytical method.
Maximum shear stress location also found by using
ANSYS Software.
REFERENCE - VIII []
(1) L.D. Rangari, P.M. Zode and P.G.
Mehar“Finite Element Analysis of LPG gas
Cylinder ”Volume 1 ,ISSUE VIII/FEB
2012PP. ISSN 2231-5063.
(2) L.D. Rangari, P.M. Zode and P.G.
Mehar“Stress Determination of LPG Gas
Cylinder by using FEM ” Vol.1, ISSUE.
X/April 2012pp.1-4 ISSN No.2231-5063.
(3) Bureu Of Indian Standard
CALCULATIONS
(4) Kaptan, A., and Kisioglu, “Finite Element
LPG Gas Cylinder IS3196
Analysis of Pressure vessel”
Let
(5) Kaptan, A., and Kisioglu, Y.,
P - Internal or burst pressure of cylinder = 2.5 Mpa
“Determination of Burst Pressures and
t - Thickness of wall cylinder = 2.5 mm.
Failure Locations of Vehicle LPG
d - Inner dia. of cylinder = 314.44 mm.
Cylinders,”.
Circumferential stress (σ)
(6) ANSYS Basic Analysis Procedure Guide
σ1= Pd/2t = 150 Mpa
Release 5.3 (1996).
Longitudinal stress
(7) Kisioglu, Y., Brevick, J. R., and Kinzel,
σ2=Pd/4t=75 Mpa
G.L., “Determination of Burst Pressure and
Maximum shear (τmax )
Locations of the DOT-39 Refrigerant
τmax= Pd/8t = 37.5 Mpa
Cylinders”, Transactions of the ASME
Equivalent Shear Stress
Journal of Pressure Vessel Technology,
= √ [ σ12- σ1 σ2- σ22] = 75 Mpa
P.P.123, 240-247, 2001.
(8) Craig, R. (1996). Mechanics of Materials.
[1]
John Wiley and Sons, New York, New
York. P.P.639
CONCLUSION
(9) Young, W. (1989). Roark’s Formulas for
Name P= 2.5 Mpa Hand
Stress & Strain. McGraw-Hill, New York,
Calcul
New York. P.P.763
ation
(10)
Max. Mpa Min. Mpa Mpa
(11)
(12)
Equivalent 153.79 0.0019614 75 (13)
stress (14)
(15)
(16) I , ISSUE - VIII [ February 2012 ]
Shear stress 86.08 0.0010221 37.5
Deformation 0.12906 0.00 0.060
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